Problem: Factor the following expression: $x^2 - 11x + 28$
Solution: When we factor a polynomial, we are basically reversing this process of multiplying linear expressions together: $ \begin{eqnarray} (x + a)(x + b) &=& xx &+& xb + ax &+& ab \\ \\ &=& x^2 &+& {(a + b)}x &+& {ab} \end{eqnarray} $ $ \begin{eqnarray} \hphantom{(x + a)(x + b) }&\hphantom{=}&\hphantom{ xx }&\hphantom{+}&\hphantom{ (a + b)x }&\hphantom{+}& \\ &=& x^2 & & {-11}x& +& {28} \end{eqnarray} $ The coefficient on the $x$ term is $-11$ and the constant term is $28$ , so to reverse the steps above, we need to find two numbers that add up to $-11$ and multiply to $28$ You can try out different factors of $28$ to see if you can find two that satisfy both conditions. If you're stuck and can't think of any, you can also rewrite the conditions as a system of equations and try solving for $a$ and $b$ $ {a} + {b} = {-11}$ $ {a} \times {b} = {28}$ The two numbers $-4$ and $-7$ satisfy both conditions: $ {-4} + {-7} = {-11} $ $ {-4} \times {-7} = {28} $ So we can factor the expression as: $(x {-4})(x {-7})$